CAT 2008 LRDI Question 11 · LR - Mathematical Reasoning

Firstly, let us try to understand the way the investments of the three traders behave.

Abdul buys shares at 10 am everyday and sells them at a particular price at 3 pm. So his return is determined by the difference in the share price at these two times.

Bikram and Chetan buy shares at equal intervals. But since Chetan buys them in equal amount he would end up buying more when the price is less and less when the price is more.

Whether the prices are continuously rising or continuously falling down or in a fluctuating market, Chetan always has a higher proportion of lower priced shares as compared to Bikram. This increases his profit in a rising market and reduces his loss in a falling market. Therefore Chetan never has return lower than that of Bikram.

We have explained this concept by taking examples. For more depth we have also provided the theoretical explanation. The theoretical explanation is only for better understanding and may not be suitable in a test environment.

Consider the scenario when the share price keeps rising throughout the day.

Let the share price at 10 am be Rs. 100, 11 am be Rs. 110, 12 noon be Rs. 140, 1 pm be Rs. 150, 2 pm be Rs. 180, and finally at 3 pm be Rs. 200.

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Abdul buys shares at Rs. 100 at 10 am and sells them at Rs. 200 at 3 pm.

∴ Abdul’s return is 100%.

Let Bikram buy one share at each interval. So, at 10 am, he buys a share for Rs. 100; at 11 am, he buys a share for Rs. 110; at 12 noon, he buys a share for Rs. 140; at 1 pm, he buys a share for Rs. 150; and at 2 pm, he buys a share for 180 × 1 = Rs. 180.

Thus, he buys a total of 5 shares for 100 + 110 + 140 + 150 + 180 = Rs. 680

At 3 pm, he sells all 5 shares for 200 × 5 = Rs. 1,000. Thus, his profit will be 1,000 − 680 = Rs. 320

Hence, Bikram's return is $\frac{320}{800}$ × 100 ≈ 47%

Let Chetan invest Rs. 415,800 at each interval. So, at 10 am, he buys 415800/100 = 4158 shares; at 11 am, he buys 415800/110 = 3780 shares; at 12 noon, he buys 415800/140 = 2970 shares; at 1 pm, he buys 415800/150 = 2772 shares; at 2 pm, he buys 415800/180 = 2310 shares.

Thus, he buys 4158 + 3780 + 2970 + 2772 + 2310 = 15990 shares for 415800 × 5 = Rs. 20,79,000. He sells these shares for 200 × 15990 = Rs. 31,98,000. His profit will be 3198000 − 2079000 = Rs. 11,19,000.

Hence Chetan's returns = $\frac{1119000}{2079000}\times 100=\frac{373}{693}$ × 100 ≈ 53%

From the above example, we see that in case of continuously rising share prices,

Abdul’s return > Chetan’s return > Bikram’s return

Thus Bikram gets the minimum return on a “boom” day.

Hence, option (a).

Note: Theoretical Explanation:

Let x1, x2, … , x6 be the share prices at 10 am, 11 am, 12 noon, 1 pm, 2 pm and 3 pm respectively.

For Abdul:

Abdul buys shares at Rs. x1 and sells them at Rs. x6.

∴ Abdul's returns = $\frac{x_6-x_1}{x_1}$

For Bikram:

Let Bikram have bought n shares at each hourly interval.

His investment amount = nx₁ + nx₂ + nx₃ + nx₄ + nx₅

= n(x₁ + x₂ + x₃ + x₄ + x₅)

$=n\times \underset{i=1}{\overset{5}{\sum x_t}}$

At 3 pm, he sells his shares for (5n × x₆)

Hence, his profit/loss = (n × 5x₆) - n × $\underset{i=1}{\overset{5}{\sum x_t}}$

= n × $\left(5x_6-\underset{i=1}{\overset{5}{\sum x_t}}\right)$

∴ Bikram's returns = $\frac{n\times (5x_6-{\displaystyle \underset{i=1}{\overset{5}{\sum x_t}}}}{n\times \underset{i=1}{\overset{5}{\sum x_t}}}=\frac{5x_6}{\underset{i=1}{\overset{5}{\sum x_t}}}-1=\frac{x_6}{{\displaystyle \frac{\underset{i=1}{\overset{5}{\sum x_t}}}{5}}}-1$

Hence, Bikram's returns = $\frac{x_6}{(\mathrm{Arithmetic} \mathrm{mean} \mathrm{of} x_1,x_2,....,x_5)}-1$

For Chetan:

Let Chetan invest Rs. P at each hourly interval.

His investment amount = 5P

Since he invests Rs. P at each interval, he buys $\frac{P}{x_1}$ shares at 10 am; $\frac{P}{x_2}$ at 11 am; and so on until 2 pm.

At 3 pm, he sells each share at x6. So, for all his shares, he receives,

Rs. $\left(\frac{P}{x_1}+\frac{P}{x_2}+\frac{P}{x_3}+\frac{P}{x_4}+\frac{P}{x_5}\right)\times x_6$

= Px₆ × $\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}\right)$

Hence, his profit/loss = Px₆ × $\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}\right)$ - 5P

= P × $\left[x_6\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}\right)-5\right]$

∴ Chetan's returns = $\frac{P\left[x_6\left({\displaystyle \frac{1}{x_1}}+{\displaystyle \frac{1}{x_2}}+{\displaystyle \frac{1}{x_3}}+{\displaystyle \frac{1}{x_4}}+{\displaystyle \frac{1}{x_5}}\right)-5\right]}{5P}$

$$\begin{gathered} =\frac{x_6\left({\displaystyle \frac{1}{x_1}}+{\displaystyle \frac{1}{x_2}}+{\displaystyle \frac{1}{x_3}}+{\displaystyle \frac{1}{x_4}}+{\displaystyle \frac{1}{x_5}}\right)-5}{5} \\ =\frac{x_6\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}\right)}{5}-1 \\ =\frac{x_6}{\left[{\displaystyle \frac{5}{\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}\right)}}\right]}-1 \end{gathered}$$

∴ Chetan's returns = $\frac{x_6}{(\mathrm{Harmonic} \mathrm{Mean} \mathrm{of} {\mathrm{x}}_1,{\mathrm{x}}_2,....,{\mathrm{x}}_5)}-1$

Now, let’s compare Bikram’s and Chetan’s returns. Since Arithmetic Mean is always greater than or equal to the Harmonic Mean, Chetan’s returns will be greater than or equal to Bikram’s.